For the disproportionation reaction
2Cu⁺(aq) ⇌ Cu(s) + Cu²⁺(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10 –1 .
Given :
(E_(Cu²⁺/Cu +)⁰ = 0.16V
E_(Cu + /Cu)⁰ = 0.52V
RT/F = 0.025)

Solution

Chemistry ma’am · 3 steps

Split the reaction into two Cu⁺ half-reactions, find E°cell, then ln K follows straight away.

1. Split the reaction. One Cu⁺ is reduced, Cu⁺ + e⁻ → Cu, at the cathode. The other is oxidised, Cu⁺ → Cu²⁺ + e⁻, at the anode. One electron moves, so n = 1.
2. Find E°cell. Both values given are reduction potentials, so E°_(cell) = E°_(cathode) - E°_(anode) = 0.52 - 0.16 = 0.36 V.

   ln K = nFE°_(cell)/RT = (1 × 0.36)/0.025 = 14.4

3. So. ln K = 14.4 = 144 × 10⁻¹, so the answer is 144.

The common slip: 288. Taking n = 2 because two Cu⁺ ions react gives ln K = 28.8, which is 288. Only one electron moves from one Cu⁺ to the other.

The idea to keep. At equilibrium Δ G = 0, so ln K = nFE°/RT. For disproportionation, the same ion is both cathode and anode, and n is the electrons moved in one balanced equation.

Answer from the NTA final answer key. Solution by PrepWiser.